Topic 10.3 Notes – Capacitors

1. What a Capacitor Is

A capacitor has two conducting surfaces separated by an insulator (a dielectric). When connected to a battery:

• One plate gets +Q
• The other gets −Q
• A potential difference $\Delta V$ builds between them

The insulator prevents charge from jumping directly across, so the charges stay separated.

Capacitance

Capacitance measures how much charge the device stores per volt:

$$C = \frac{Q}{\Delta V}$$

• Units: farads (F) = C/V
• $C$ depends only on geometry and materials, not on $Q$ or $\Delta V$

Think of capacitance as a built-in property of the setup, like resistance for a resistor.

2. Geometries of Capacitors

AP Physics C only expects three shapes. All of them follow the same strategy:

1. Use Gauss’s law to find $E(r)$
2. Integrate to find $\Delta V$
3. Use $C = Q/\Delta V$

Parallel-Plate Capacitor

Two large plates, area $A$, separation $d$, connected to a battery that maintains a potential difference.

Parallel-plate capacitor connected to a battery

Capacitance:

$$C = \kappa \varepsilon_0 \frac{A}{d}$$

• $A \uparrow \Rightarrow C \uparrow$
• $d \uparrow \Rightarrow C \downarrow$
• $\kappa \uparrow \Rightarrow C \uparrow$

Here $\kappa$ is the dielectric constant and $\varepsilon_0$ is the permittivity of free space.

This formula is heavily tested. You should instantly see how changing geometry affects $C$.

Concentric Spherical Capacitor

Two conducting spheres, inner radius $a$, outer radius $b$.

$$C = 4\pi \varepsilon_0 \frac{ab}{b-a}$$

Special case: if the outer sphere is very far away, it behaves like an isolated sphere:

$$C = 4\pi \varepsilon_0 R$$

Students often forget this limit. It shows up in conceptual questions.

Coaxial Cylindrical Capacitor

Two long cylinders, inner radius $a$, outer radius $b$, length $L$.

$$C = \frac{2\pi \varepsilon_0 L}{\ln(b/a)}$$

Notice the logarithm. That comes from integrating $1/r$ when using Gauss’s law.

3. Electric Field Between Parallel Plates

For large plates where $d$ is much smaller than plate dimensions, the field is uniform between them.

Using Gauss’s law:

• Surface charge density $\sigma = Q/A$
• Field from one sheet: $E = \sigma / (2\varepsilon_0)$
• Between two plates (fields add):

$$E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$$

Key connection:

$$E = \frac{\Delta V}{d}$$

Combining these gives $C = \varepsilon_0 A/d$ (no dielectric).

With a dielectric inserted:

$$E = \frac{\sigma}{\kappa \varepsilon_0}$$

So the dielectric reduces the field and increases capacitance.

Charged Particle Between Plates

Uniform field → constant force:

$$F = qE \quad \Rightarrow \quad a = \frac{qE}{m}$$

That means the motion is mathematically identical to projectile motion.

If a problem mixes horizontal velocity with vertical electric acceleration, treat it exactly like projectile motion.

4. Energy Stored in a Capacitor

As charge builds up, the voltage increases linearly. The average voltage during charging is $\frac{1}{2}\Delta V$.

Energy stored:

$$U = \frac{1}{2} Q \Delta V$$

Equivalent forms:

$$U = \frac{1}{2} C (\Delta V)^2$$

$$U = \frac{Q^2}{2C}$$

Pick the version that matches what you’re given so you avoid extra algebra.

Energy Density

For parallel plates, the energy is stored in the electric field:

$$u = \frac{1}{2} \varepsilon_0 E^2$$

With dielectric: $u = \frac{1}{2} \kappa \varepsilon_0 E^2$.

This idea is important on conceptual questions. The energy is in the field between the plates, not sitting on the charges themselves.